Universitext
Editorial Board
(North America):
S. Axler
K.A. Ribet
Falko Lorenz
Algebra
Volume I:
Fields and Galois Theory
With the collaboration of the translator, Silvio Levy
www.pdfgrip.com
Falko Lorenz
FB Mathematik Institute
University Muănster
Muănster, 48149
Germany
Editorial Board
(North America):
K.A. Ribet
Mathematics Department
University of California at Berkeley
Berkeley, CA 94720-3840
USA
S. Axler
Mathematics Department
San Francisco State University
San Francisco, CA 94132
USA
Mathematics Subject Classification (2000): 11-01, 12-01, 13-01
Library of Congress Control Number: 2005932557
ISBN-10: 0-387-28930-5
ISBN-13: 978-0387-28930-4
Printed on acid-free paper.
© 2006 Springer Science+Business Media, Inc.
All rights reserved. This work may not be translated or copied in whole or in part without the
written permission of the publisher (Springer Science+Business Media, Inc., 233 Spring Street, New
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“Though what I’m saying is perhaps not new, I have
felt it quite vividly on this new occasion.”
J. W. Goethe, in a letter from Naples, 17 May 1787
Foreword
The present textbook is my best effort to write a lively, problem-oriented and understandable introduction to classical modern algebra. Besides careful exposition, my
goals were to lead the reader right away to interesting subject matter and to assume
no more background than that provided by a first course in linear algebra.
In keeping with these goals, the exposition is by and large geared toward certain
motivating problems; relevant conceptual tools are introduced gradually as needed.
This way of doing things seems more likely to hold the reader’s attention than a
more or less systematic stringing together of theorems and proofs. The pace is more
leisurely and gentle in the beginning, later faster and less cautious, so the book lends
itself to self-study.
This first volume, primarily about fields and Galois theory, in order to deal with
the latter introduces just the necessary amount of group theory. It also covers basic
applications to number theory, ring extensions and algebraic geometry. I have found
it advantageous for various reasons to bring into play early on the notion of the
algebraic closure of a field. Naturally, Galois’ beautiful results on solvable groups
of prime degree could not be left out, nor could Dedekind’s Galois-theoretical arithmetic reduction principle. Infinite Galois extensions are not neglected either. Finally,
it seemed appropriate to include the fundamentals of transcendental extensions.
At the end of the volume there is a collection of exercises, interspersed with
remarks that enrich the text. The problems chosen are of widely varying degrees of
difficulty, but very many of them are accompanied by hints — sometimes amounting
to an outline of the solution — and in any case there are no outright riddles. These
exercises are of course meant to allow readers to practice their grasp of the material,
but they serve another important purpose as well: precisely because the main text
was kept short and to the point, without lots of side-results, the appendix will give
the reader a better idea of the wealth of consequences and applications derived from
the theory.
The linear algebra facts used, when not totally elementary, are accompanied by
references to my Lineare Algebra, now published by Spektrum Akademischer Verlag
and abbreviated LA I and LA II. This has not been translated, but equivalent spots in
other linear algebra textbooks are not hard to find. Theorems and lesser results are
numbered within each chapter in sequence, the latter being marked F1, F2, : : : — the
F is inherited from the German word Feststellung. Allusions to historical matters
are made only infrequently (but certainly not at random). When a theorem or other
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vi
Foreword
result bears the name of a mathematician, this is sometimes a matter of tradition
more than of accurate historical origination.
The first German edition of this book appeared in 1987. I thank my colleagues
who, already back at the writing stage, favored it with their interest and gave me
encouragement — none more than the late H.-J. Nastold, with whom I had many
fruitful conversations, W. Lütkebohmert, who once remarked that there was no
suitable textbook for the German Algebra I course, O. Willhöft, who suggested
several good problems, and H. Schulze-Relau and H. Epkenhans, whose critical
perusal of large portions of the manuscript was a great help. The second (1991) and
third (1995) editions benefited from the remarks of numerous readers, to whom I
am likewise thankful, in particular R. Alfes, H. Coers, H. Daldrop and R. Schopohl.
The response and comments on the part of students were also highly motivating.
Special thanks are due to the publisher BI-Wissenschaftsverlag (later acquired by
Spektrum) and its editor H. Engesser, who got me going in the first place.
The publication of this English version gives me great pleasure. I’m grateful
to Springer-Verlag New York and its mathematics editor Mark Spencer, for their
support and competent handling of the project. And not least for seeing to it that
the translation be done by Silvio Levy: I have observed the progress of his task with
increasing appreciation and have incorporated many of the changes he suggested, in
a process of collaboration that led to noticeable improvements. Further perfecting
is of course possible, and readers’ suggestions and criticism will continue to be
welcome and relevant for future reprints.
Münster, July 2005
Falko Lorenz
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Contents
Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
1
Constructibility with Ruler and Compass . . . . . . . . . . . . . . . . . . . . . . . . .
1
2
Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3
Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
4
Fundamentals of Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
5
Prime Factorization in Polynomial Rings. Gauss’s Theorem . . . . . . . . .
45
6
Polynomial Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
7
Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
8
Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
9
Finite Fields, Cyclic Groups and Roots of Unity . . . . . . . . . . . . . . . . . . .
83
10
Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
11
Applications of Galois Theory to Cyclotomic Fields . . . . . . . . . . . . . . . . 103
12
Further Steps into Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
13
Norm and Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
14
Binomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
15
Solvability of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
16
Integral Ring Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
17
The Transcendence of
18
Transcendental Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
19
Hilbert’s Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
Appendix: Problems and Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
Index of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
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1
Constructibility with Ruler and Compass
1. In school one sometimes learns to solve problems where a certain geometric
figure must be constructed from given data. Such construction problems can be
quite difficult and afford a real challenge to the student’s intelligence and ingenuity.
If you have tried long and hard to solve a certain construction problem, to no
avail, you might then wonder whether the required construction can be carried out at
all. Whether a construction exists is a fundamental question: so much so that there
are some construction problems that had already been entertained by the ancient
Greeks, and yet remained unsolved for two thousand years and more.
For example, nobody has ever been able to state a procedure capable of dividing
an arbitrary angle into three equal angles, using ruler and compass. Now, of course
construction problems range widely in degree of difficulty: think of the similarlooking problem of dividing an arbitrary segment into three equal parts using ruler
and compass — not totally trivial, but after some thought just about anyone can
carry out the construction. The problem of constructing a regular pentagon is also
solvable, but already somewhat more complicated. So it is certainly understandable
that even a construction problem that has eluded would-be solvers for a long time
should leave room for hoping that success might yet be achieved through greater
ingenuity. Perhaps, then, the question of whether a particular construction with ruler
and compass is possible is not one that comes to mind immediately.
Even if someone asks this question of principle, it is not clear a priori that there
is a promising way to tackle it. Yet there is, as the development of algebra since
Gauss (1777–1855) has shown. I would like to explain now, at the beginning of our
introduction to algebra, how one can arrive at broad statements about the general
constructibility problem, by translating this geometric problem into an algebraic one.
As we elaborate on this, we will have the chance to motivate quite naturally certain
fundamental algebraic concepts. Moreover the subsequent treatment of the derived
algebraic problem will require many of the tools usually treated in an Algebra I
course. This procedure has the advantage that one starts from a concrete and easily
understood question and keeps the goal of solving it in mind as one goes along.
Let it be said, however, that the problem of constructibility with ruler and compass by no means played a central role in the development of algebra. In this
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1 Constructibility with Ruler and Compass
regard the problem of solving algebraic equations by means of radicals was surely
more significant, not to mention other motivations and stimuli coming from outside
algebra — from number theory and analysis, for example. Incidentally, in due time
we will make precise the problem of solubility of equations by radicals and keep it
in view as the exposition unfolds.
2. First let’s describe properly what is to be understood by constructibility with
ruler and compass. For this we start from the plane ޒ2 of elementary geometry. A
construction problem asks whether a certain point P of the plane can be constructed
with ruler and compass, starting from a given initial set M of points. Thus, let a
subset M of ޒ2 be given (we may as well assume it has at least two points). Then
look at the set
ˇ
«
˚
M D P 2 ޒ2 ˇ P is constructible from M with ruler and compass ;
to be defined more precisely as follows. Let
Li.M / D set of straight lines joining two distinct points of M ,
Ci.M / D set of circles whose center belongs to M
and whose radius equals the distance between two points of M .
Then consider the following elementary steps for the construction of “new” points:
(i) intersecting two distinct lines in Li.M /;
(ii) intersecting a line in Li.M / with a circle in Ci.M /;
(iii) intersecting two distinct circles in Ci.M /.
Let M 0 be the union of M with the set of points obtained by the application of
one of these steps. The points of ޒ2 that can be obtained by repeated application
of steps (i)–(iii), starting from M and replacing M by M 0 each time, are said to be
constructible from M with ruler and compass. They form the set M .
We just mention right now four well known constructibility problems that were
posed already by the ancient Greeks.
Q0
˛: Trisection of the angle.
Given an angle of measure ', construct an angle of
measure '=3 with ruler and compass. We regard
the given angle as determined by its vertex S and
points Q; Q0 on each of its sides; one may as well
assume that Q and Q0 are equidistant from S . Let
X be the point indicated in the figure. The question
then is whether X 2 fS; Q; Q0 g.
X
'
S
'=3
Q
ˇ: Doubling of the cube (Delian problem).
Given a cube of side length a, find a cube of twice
the volume. The side length x of
p
3
the desired cube satisfies x 3 D 2a3 , so x p
D a 2. Thus let P; Q; X be points on the
3
real line such that PQ D a and PX D a 2; the question is whether X 2 fP; Qg.
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Classical construction problems
3
: Quadrature of the circle.
Given a circle, construct a square of the same surface area. The given circle of
radius r is determined by points P and Q a distance r D
pPQ apart. The side length
x of the desired square must satisfy x 2 D r 2 , so x
D
r
. What must be decided,
p
then, is whether some point X such that PX D r
belongs to fP; Qg.
ı: Construction of a regular n-gon (n-section of the circle).
As before, we think of the circle as being given by
two points P and Q, the center and a point on the
circumference. Let X be the point shown on the
right. For what natural numbers n does X lie in
fP; Qg ?
This is the case, for example, for n D 6 (and
therefore for all numbers of the form n D 3 2m ):
it is enough to draw a circle of radius PQ with
center Q. By successively repeating the procedure
with the newly found points one gets the well known rosette:
X
2 =n
P
Q
3. To make the problem of constructibility with ruler and compass accessible, one
must first “algebraize” it. To that end it is useful to employ the identification
ޒ2 D ;ރ
that is, to regard points in the plane as complex numbers, and so take advantage of
the possibility not only of (vector) addition but also of multiplication. Assuming
the basic properties of the field ރof complex numbers, the problem of dividing the
circle into n parts (see ı above) amounts to the following question: Is it the case
that
e 2 i=n 2 f0; 1g ?
The next statement points out that the fundamental algebraic operations of ރ
can be described constructively.
F1. Let M be any subset of ރcontaining the numbers 0 and 1. Then:
(1) i 2 M ;
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4
1 Constructibility with Ruler and Compass
(2) z 2 M ) z 2 M ;
(3) z 2 M ) Re z; Im z 2 M ;
(4) z 2 M ) z 2 M ;
(5) z1 ; z2 2 M ) z1 C z2 2 M ;
(6) z1 ; z2 2 M ) z1 z2 2 M ;
(7) z 2 M; z ¤ 0 ) 1=z 2 M .
Proof. (1) The line connecting 0 and 1, that is, the real line ޒ, belongs to Li.M / by
definition. Intersecting ޒwith the unit circle, which belongs to Ci.M /, we see that
1 2 M . If we now construct the perpendicular bisector of the interval Π1; 1 in
the well-known way and intersect it with the unit circle, we obtain i 2 M .
(2) Drop a perpendicular from z to ޒ. From the foot of this perpendicular, say a,
draw a circle whose radius is the distance from a to z. Its second intersection with
the straight line through z and a gives z 2 M .
(3) As just verified, we have a D Re z 2 M . To obtain b D Im z, draw the
perpendicular to the imaginary axis through z, and then transfer to ޒthe absolute
value of the foot bi of the perpendicular.
(4) Intersect the line through 0 and z with the circle of radius jzj and center 0.
(5) Intersect the circle of center z1 and radius jz2 j with the circle of center z2 and
radius jz1 j. One of the intersections is the vertex z1 C z2 of the parallelogram
determined by z1 ; z2 .
(6) If z1 D a1 C i b1 and z2 D a2 C i b2 we have
z1 z2 D .a1 a2
b1 b2 / C .a1 b2 C a2 b1 /i:
Now z1 ; z2 2 M implies a1 ; b1 ; a2 ; b2 2 M , by (3). If we assume the claim is
true for real numbers, it will also be true for arbitrary complex numbers, because of
(4) and (5). Therefore we must prove that given real numbers r1 and r2 ,
r1 ; r2 2 M ) r1 r2 2 M:
Clearly one can assume r1 ; r2 > 0. To complete the proof, consider this diagram:
z
0
1
r1
r2
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x
Algebraization of the problem
5
Here z is the appropriate intersection of the line through 0 and 1 C i with the
circle of radius r2 and center 0, and the dashed line indicates a parallel to the line
through 1 and z. By similarity of triangles we have xWr2 D r1 W1, and therefore
x D r1 r2 . Since x lies in M , this proves the claim.
(7) Since z 1 D z .zz/ 1 , it suffices in view of the earlier parts to show that if
r > 0 lies in M , so does r 1 . To do this we refer to the following diagram:
0
x
1
r
By similarity of triangles, r W1 D 1Wx, and this proves the claim since x 2 M .
˜
As a consequence of F1 we will explicitly state again:
F2. Let M be a subset of ރcontaining the points 0 and 1. Then
ރ. It is called the field of numbers constructible from M .
M is a subfield of
In particular, ޑÂ M , since ޑis the smallest subfield of ރ. Also the set
fa C bi j a; b 2 ޑg — which incidentally is also a subfield of M — is contained
in M . But the field M is substantially larger:
F3. The field M is quadratically closed, that is, for every z 2 ރwe have
p
(8)
z 2 M ) z 2 M;
p
where z represents any complex number w with w 2 D z .
p
Proof. Suppose p
w 2 D z D r e i' . Letting r be the positive square root of r 2 ޒ,
we have w D ˙ r e i'=2 . Since it is always possible to bisect an angle with ruler
and compass, it p
is enough in order to prove (8) to show that for any r > 0 in M ,
the square root r is also in M . To do this we raise the perpendicular to the
segment Π1; r through 0 and intersect it with the semicircle constructed over the
same segment, to obtain a point v:
v
1
0
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r
6
1 Constructibility with Ruler and Compass
Then x D jvj belongs to
M . By Thales’ Theorem the triangle with vertices
1; v; r has a right angle; applying
the formula for the altitude of a right triangle
p
we get x 2 D 1 r , and so x D r .
˜
4. Now let’s return to the statement of F2, which says that M is a field containing
ޑas a subfield. Trivially,
M Â M:
(9)
Thus the field E WD M also contains all numbers that can be obtained from k WD ޑ
and from M by means of arithmetic operations in E. This serves as the motivation
for a simple but fundamental definition:
Definition 1. Let E be a field and k a subfield of E. We say that E is an extension
of k. Let A be any subset of E. Set
k.A/ WD intersection of all subfields F of E such that k  F and A  F .
We call k.A/ the subfield of E generated by A over k, and we also say that k.A/
arises from k by adjoining to k the elements of A. (The corresponding noun is
adjunction.) Clearly, k.A/ is the smallest subfield of E containing k and A. In the
case of a finite set A D f˛1 ; : : : ; ˛m g we also denote the field k.A/ by
k.˛1 ; ˛2 ; : : : ; ˛m /:
Example. Take E D ރ, k D ޑ, A D fi g. We claim that
ޑ.i / D fa C bi j a; b 2 ޑg:
Proof. Let F0 be the set on the right-hand side. Then ޑ F0 and fi g  F0 no
matter what. Since ޑ.i / is a subfield of E, we also have F0 Â ޑ.i /. To prove the
claim we must show that F0 is a subfield of ރ. Clearly, F0 is closed under addition
and multiplication. There remains to show that if z D a C bi Ô 0 lies in F0 , so does
z 1 . But
a ib
a
b
D 2
i 2
;
z 1 D z.zz/ 1 D 2
2
2
a Cb
a Cb
a Cb 2
so z
1
˜
does lie in F0 .
Warning. In the situation of Definition 1 it is not generally true that
k.˛/ D fa0 C a1 ˛ C
C an ˛ n j ai 2 k; n
0g:
We will return to this point in Chapter 3.
Now let’s return to the earlier situation. We know that
M is an extension of
ޑcontaining M . Thus it contains the well-defined subfield ޑ.M / obtained from
ޑby adjunction of M . We set M D fz j z 2 M g and consider the subfield
(10)
K WD ޑ.M [ M /
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Algebraization of the problem
7
of M obtained from ޑby adjunction of the set A WD M [ M (because of (2) and
(9), we know that A
M ). Since M Â K Â M we obviously have
M D K:
(11)
In other words, when considering the set of numbers constructible from a set, the
initial set M can always be replaced by the field given by (10). By the way, the
field K is mapped to itself under complex conjugation:
K D K:
(12)
This is clear from (10), but we will justify it in detail for pedagogical purposes: By
definition, K D ޑ.A/, and so ޑD ޑÂ K and A D A Â K. By Definition 1, then,
ޑ.A/ Â K, that is, K Â K. By complex conjugation we then get K Â K D K.
(Where have we used the fact that the complex conjugate of a subfield of ރis also
one?)
F4. Let K be any subfield of ރwith K D K.
(a) If z is the intersection of two distinct lines in Li.K/, then z 2 K.
(b) If z is an intersection of a line in Li.K/ with a circle in Ci.K/, then
. /
there exists w 2 ރwith w 2 2 K and z 2 K.w/.
(c) If z is an intersection of two distinct circles in Ci.K/, condition . / again holds.
These statements simply reflect the well-known fact that the analytic counterpart of the elementary construction steps (i)–(iii) can only lead to solving linear
or quadratic equations. Therefore we postpone the proof of F4 and instead derive
from F4 certain consequences of great import to the constructibility problem. First
we equip ourselves with appropriate terminology:
Definition 2. Let E be an extension of the field K.
(a) We say that E arises from K by adjoining a square root if there exists w 2 E
such that
w 2 2 K and E D K.w/:
p
We call w a square root of the element v WD w 2 of K, and we write w D v.
(b) We say that E arises from K by successively adjoining square roots if there
is a chain K D K0 Â K1 Â
 Km D E of subfields Ki of E where each
Ki is obtained from Ki 1 by adjoining a square root.
p
Examples. 1. E D ޑ. 2/ is obtained from ޑby adjoining a square root.
2 i=3
2. E D ޑ.z/, where z D ep
, is obtained from ޑpby adjoining a square root.
1
1
For since z D 2 C 2 i 3, we have ޑ.z/ D ޑ.
3/.
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8
1 Constructibility with Ruler and Compass
3. E D ޑ.e 2 i=5 / is obtained from ޑby successively adjoining square roots.
This is not immediately obvious, but can be seen as follows: The complex
number w D e 2 i=5 satisfies w 4 C w 3 C w 2 C w C 1 D 0, since w 5 D 1 and
w ¤ 1. Dividing by w 2 yields
w2 C w1 C 1 C w
(13)
Now set z WD w C w
1
1
Cw
2
D 0:
; then (13) becomes, via .w C w
/ D w2 C 2 C w
1 2
2
,
z2 C z
1 D 0:
p
This equation has the solutions z D 12 ˙ 12 5. The field K1 WD ޑ.z/ therefore
satisfies
p
K1 D ޑ. 5/:
(14)
But E D K1 .w/, and w obviously satisfies the quadratic equation
(15)
w2
zw C 1 D 0;
whose coefficients lie in K1 . Thus, as can be seen from the quadratic formula,
E is obtained from K1 by adjoining a square root of z 2 4, which lies in K1 .
The following theorem fulfills our first goal, the reduction of the geometric
problem of constructibility with ruler and compass to a purely algebraic problem.
Theorem 1. Suppose M Â ރcontains 0 and 1. Set
K WD ޑ.M [ M /:
For a given z 2 ރ, the following statements are equivalent:
(i) z 2 M , that is, z is constructible from M with ruler and compass.
(ii) z lies in a subfield E of ރobtainable from K by successively adjoining square
roots.
Proof. (ii) ) (i): By assumption, there exists a finite chain of subfields of ރ, say
K D K0 Â K1 Â
 Km D E;
satisfying Ki D Ki 1 .wi / with wi2 2 Ki 1 for each i , and also z 2 E. We also
know that K0 D K Â M . Now consider the field K1 D K.w1 /, with w12 2 K. By
F3, w1 lies in M because w12 does. Since M is a field, we have K.w1 / Â M ,
that is, K1 Â M . Analogously we get K2 Â M , and so on until we finally get
E D Km  M . Since z 2 E we have z 2 M .
(i) ) (ii): We first consider a z 2 ރarising from M by applying only one of the
elementary construction steps (i), (ii), (iii). Now we make use of F4. We conclude
that z 2 K in the case of step (i); in cases (ii) and (iii) we get z 2 K.w/, where
w 2 ރis such that w 2 lies in K. We claim that in each case z lies in a subfield
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Algebraization of the problem
9
K 0 of ރobtained from K by successively adjoining square roots and satisfying the
condition
(16)
K0 D K0:
In case (i) this is clear; see (12). In cases (ii) and (iii) consider the field
(17)
K 0 D K.w; w/ D K.w/.w/:
It clearly satisfies condition (16). Since w 2 2 K, we have w 2 D w 2 2 K D K Â K.w/,
so by (17) K 0 is indeed obtained from K by successively adjoining square roots.
For an arbitrary z 2 M the assertion follows by induction on the number of
elementary steps needed to construct z.
˜
Now that Theorem 1 has accomplished the desired algebraization of our problem,
the four classical constructibility problems listed earlier can also be reformulated
algebraically:
p
3
Doubling of the cube: Does 2 lie in a subfield of ރobtainable from ޑthrough
the successive adjunction of square roots?
Quadrature of the circle: Does the number lie in a subfield of ރthat can be
obtained from ޑby successively adjoining square roots?
Construction of a regular n-gon: For what natural numbers n is the complex
number e 2 i=n contained in a subfield of ރobtainable from ޑby the successive
adjunction of square roots? (By Example 3 after Definition 2, this is certainly the
case for n D 5: thus a regular pentagon is constructible with ruler and compass. You
are encouraged to derive a practical construction from the calculation in Example 3;
it is not hard to do.)
Angle trisection: Let ' be any real number. Is it the case that the complex
number e i'=3 always lies in a subfield of ރobtained from the field
(18)
K D ޑ.e i' /
by successively adjoining square roots? (Note: For z WD e i' we have z D e i' D z 1 ,
so ޑ.z; z/ D ޑ.z/, and (18) does indeed represent the right ground field for the
purposes of Theorem 1.)
5. We now carry out the proof of F4. We start with an arbitrary subfield K of ރ
satisfying K D K.
(a) An arbitrary line g in ޒ2 D ރis given by an equation
(19)
g D fz0 C t z1 j t 2 ޒg;
where z0 ; z1 2 and z1 Ô 0. If g 2 Li.K/ we can assume that z0 ; z1 2 K. Now
suppose that g0 D fz00 C t 0 z10 j t 0 2 ޒg, with z00 ; z10 2 K, is another line in Li.K/,
distinct from g, and that z 2 g \ g0 . There exist uniquely determined real numbers
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10
1 Constructibility with Ruler and Compass
t; t 0 such that z D z0 C t z1 D z00 C t 0 z10 . Looking separately at real and imaginary
parts, we see that .t; t 0 / is the unique solution of the system of linear equations
tx1
t 0 x10 D x00
x0 ;
ty1
t 0 y10
y0 :
D
y00
Multiplying the second equation by i we get a system of linear equations over K.
It follows that t; t 0 2 K and so z 2 K.
(b) An arbitrary circle c in ޒ2 D ރis given by
˚
«
(20)
c D z j .z a/.z a/ D r 2 ;
with a 2 ރand r > 0 real. Suppose c 2 Ci.K/. Then a 2 K. Since c also contains
an element b from K, and since r 2 D .b a/.b a/ and K D K, we have r 2 2 K.
Now let g 2 Li.K/ be given by (19) and suppose z 2 g \ c. Then there exists t 2 ޒ
with z D z0 C t z1 and, in view of (20),
.z0 C z1 t
a/.z 0 C z 1 t
a/ D r 2 :
Multiplying out and dividing by jz1 j2 , we get an equation
t 2 C pt C q D 0;
with p and q in K. Then w WD t C 12 p satisfies w 2 2 K, and since z D z0 C t z1 and
K.t / D K.w/ we have z 2 K.w/.
(c) Now let two distinct circles c1 and c2 be given and suppose z 2 c1 \ c2 . Then z
satisfies a system of equations of the form
(21)
.z
a/.z
a/ D r 2 ;
.z
b/.z
b/ D s 2 ;
with a; b; r 2 ; s 2 2 K and a Ô b. Subtracting one equation from the other yields
(22)
z.b
a/ C z.b
a/ D c;
with c D r 2 s 2 aa C bb 2 K. Solving equation (22) for z and substituting the
resulting value into the first line of (21), we get for z a quadratic equation with
coefficients in K. The assertion follows immediately.
˜
6. The algebraic translation of the constructibility problem (Theorem 1) thus leads
us to a more detailed study of the extensions of a given field K. In this forthcoming
investigation the following statement is both simple and fundamental:
F5 (Dedekind). Let K be a field and E an extension of K. Then E can be regarded
as a vector space over K.
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The degree of a field extension
11
Proof. This is clear: we consider on E the existing addition operation and a scalar
multiplication K E ! E defined simply by restricting the existing multiplication
map E E ! E of the field E. With these operations E of course obeys all the
axioms of a K-vector space. (Incidentally, by considering on the K-vector space
E thus obtained the original multiplication operation, we can make E into a Kalgebra; for this concept see LA I, p. 87.)
˜
By regarding extensions of a field K as K-vector spaces, one gains access to
the powerful methods of linear algebra, which have demonstrated their fruitfulness
over and over in many different areas of mathematics and applications.
Definition 3. If E is an extension of the field K, we denote by
ŒE W K
the dimension of the K-vector space E. Instead of ŒE W K we can also write E W K.
This number is called the degree of E over K.
Examples. (1) As an ޒ-vector space, ރD ޒ2 , so ރW ޒD 2.
(2) We have ޑ.i / W ޑD 2; see the example after Definition 1 and observe that
i … ޑ.
p
3
(3) We will see in Chapter 2 that ޑ. 2/ W ޑD 3.
(4) Because ޒis uncountable, the degree ޒW ޑcannot be finite.
The usefulness of the viewpoint introduced in F5 becomes apparent already from
the next statement:
F6. Let E be an extension of the field K, and suppose that 1 C 1 Ô 0 in K. Then
these two statements are equivalent:
(i) E W K D 2.
(ii) E is obtained from K by adjoining a square root that is not already in K.
Proof. Suppose (i) holds, and let ˛ be an element of E not belonging to K. Since
E W K D 2, the set f1; ˛g is necessarily a K-basis of E. In particular there is a
relation of the form
˛ 2 C p˛ C q1 D 0;
with p; q 2 K:
For w WD ˛ C 12 p we then have w 2 D 14 p 2 q 2 K. Since E D K.˛/ D K.w/ this
implies (ii).
Suppose, conversely, that E D K.w/ with w 2 DW d 2 K and w … K. Clearly
0
E WD fa C bw j a; b 2 Kg is a subring of E containing K. To prove (i) therefore
we just have to show that for every a C bw Ô 0 in E 0 the inverse .a C bw/ 1 also
belongs to E 0 . This follows from
.a C bw/.a bw/ D a2 b 2 d 2 K;
p
because we know (from w D d … K) that a2 b 2 d ¤ 0.
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˜
12
1 Constructibility with Ruler and Compass
Thanks to F6 we can recast Theorem 1 as follows:
Theorem 10 . As before, suppose M Â ރcontains 0 and 1, and set K WD ޑ.M [M /.
Then these two statements are equivalent:
(i) z 2 M .
 Km of subfields of ރsuch that
(ii) There is a finite chain K D K0 Â K1 Â
z 2 Km and
Ki W Ki 1 D 2 for 1 Ä i Ä m:
This result suggests that we should study the relationship between the degrees
of the extensions in the diagram
(23)
E
ˇ
ˇ
ˇ
ˇ
Fˇ
ˇ
ˇ
ˇ
K
whose meaning is that F is an extension of K and E is an extension of F .
F7 (Degree formula). Let E be an extension of K and let F be a subfield of E
containing K. Then
(24)
ŒE W K D ŒE W F ŒF W K:
Proof. If E W K is finite, so are E W F and F W K, of course. Now assume
(25)
F W K D m and E W F D n;
with m and n natural numbers. We show that E W K is also finite and satisfies
equation (24). Indeed, by (25) there is an isomorphism F ' K m of K-vector spaces
and an isomorphism E ' F n of F -vector spaces. This results in an isomorphism
E ' F n ' .K m /n D K mn
of K-vector spaces. It follows that E W K D mn, so equation (24) holds. The essential
content of (24) is thus proved. (Incidentally, it is clear how to modify the argument
in case any of the degrees are infinite, so as to prove (24) regarded as an equality
between cardinals.) But in addition we establish the following: If ˛1 ; : : : ; ˛m form
a basis of F over K and ˇ1 ; : : : ; ˇn form a basis of E over F , the elements
(26)
.˛i ˇj /1ÄiÄm; 1Äj Än
formPa basis of the K-vector space E. For any ˛ 2 E can be written in P
the form
˛ D j bj ˇj with coefficients bj P
2 F ,P
which in turn can
be
written
as
b
D
j
i aij ˛i
P
with aij 2 K; it follows that ˛ D j
a
˛
D
a
˛
ˇ
,
so
the
elements
ˇ
ij
i
j
ij
i
j
i
i;j
in (26) span the K-vector space E. Since E W K D mn they must form a basis. ˜
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The degree of a field extension
13
We immediately get, as a consequence of F7 (and F6):
F8. If E is obtained from K by successively adjoining square roots,
E W K D 2m
for some integer m
0.
Remark. The converse of F8 is unfortunately not true; see Appendix, §5.7.
At any rate, F8 gives us a necessary condition for a number to be constructible:
F9. Let K be a subfield of ރwith K D K. If z 2 ރis constructible from K,
(27)
K.z/ W K
is a power of 2:
Proof. Take z 2 K. By Theorem 1, z lies in an extension E of K that can be
obtained from K by successively adjoining square roots. By F8, E W K D 2m is
a power of 2. Since z 2 E we have K.z/ Â E. Because of the degree formula
ŒE W K D ŒE W K.z/ ŒK.z/ W K, the integer K.z/ W K is a power of 2, since it divides
E W K.
˜
As remarked, the converse of F9 is not generally true. Only in Chapter 11 will
we be able to explain how condition (27) can be modified to give a necessary and
sufficient condition for the constructibility of a number.
Regarding the four classical constructibility problems listed near the beginning
of this chapter, F9 tells us that we should be investigating the following questions:
˛/ ޑ.e i'=3 / W ޑ.e i' / D ?
p
3
ˇ/ ޑ. 2/ W ޑD ?
/ ޑ. / W ޑD ?
ı/ ޑ.e 2 i=n / W ޑD ?
p
3
If we can show, for example, that ޑ. 2/ W ޑD 3 , this would prove that the
problem of the doubling of the cube is insoluble with ruler and compass.
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2
Algebraic Extensions
1. Let K be a field and E an extension of K. One writes this assumption in short
as
Let E=K be a field extension;
and the word “field” is often omitted when it can be inferred from the context.
An element ˛ of E is called algebraic over K if there exists a polynomial
f .X / Ô 0 in KX such that
f .˛/ D 0:
If ˛ is not algebraic over K, we say that ˛ is transcendental over K.
Remarks. (a) If K D ޑand E D ރ, the elements of E algebraic over K are called
simply algebraic numbers, and the elements of Eptranscendental over K are
3
called transcendental numbers. Example: ˛ WD 2 is an algebraic number,
3
since ˛ is a root of the polynomial X
2 2 ޑŒX .
(b) The set of algebraic numbers is countable (since ޑŒX is countable and any
nonzero polynomial in ޑŒX has finitely many roots in ) ރ. Therefore the set
of transcendental numbers must be uncountable. To actually be able to exhibit
a transcendental number is a different (and much harder) matter.
Theorem 1. Let M be a subset of ރcontaining 0 and 1. Any point z 2
algebraic over K WD ޑ.M [ M /.
M is
The proof will be given later in this chapter. But first we quote a famous result:
Theorem 2 (Lindemann 1882). The number
is transcendental.
Corollary. The quadrature of the circle with ruler and compass is impossible.
Proof. If it were possible, we would have 2 ; ޑby Theorem 1 then
algebraic, which by Lindemann’s Theorem is not the case.
would be
˜
Lindemann’s Theorem can be proved using relatively elementary algebraic and
analytic arguments, but the proof is on the whole quite intricate. We will go into it
later on (Chapter 17).
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16
2 Algebraic Extensions
2. Now we start our study of field theory with the following statement:
F1. Let E=K be a field extension. If ˛ 2 E is algebraic over K, then
K.˛/ W K < 1:
Proof. Suppose there exists a nonzero polynomial
(1)
f .X / D X n C an
1X
n 1
C
C a0 2 KŒX
such that f .˛/ D 0; we have assumed without loss of generality that f is normalized
(has leading coefficient 1). There exists a unique homomorphism of K-algebras '
from the polynomial ring KŒX into E such that '.X / D ˛ (see page 21); its image
R D im '
E
consists precisely of those elements of E that can be written as polynomial expressions g.˛/ in ˛ with coefficients in K. But in writing such an expression we
immediately see from the relation
(2)
˛ n D .an
1˛
n 1
C
C a1 ˛ C a0 /
that only terms of degree less than n are needed, so in fact
(3)
R D fc0 C c1 ˛ C
C cn
1˛
n 1
j ci 2 Kg:
Thus, as a vector space over K, the dimension of R is at most n. Since R, being
a subring of E, has no zero-divisors, a simple argument (given a bit further down)
shows that R is actually a field. It follows that K.˛/ Â R (using the definition of
K.˛/), and therefore that R D K.˛/. From (3) we then get
(4)
K.˛/ D fc0 C c1 ˛ C
C cn
1˛
n 1
j ci 2 Kg:
In particular,
(5)
K.˛/ W K Ä n.
F2. Let R be an integral domain (that is, a commutative ring with no zero divisors
and with 1 Ô 0), and let K be a subfield of R. If R is finite-dimensional as a K-vector
space, R is a eld.
Proof. For a given a Ô 0 in R, consider the map h W R ! R given by multiplication
by a, namely, h.x/ D ax for all x in R. Then h is an endomorphism (linear map)
of the K-vector space R. Since R has no zero-divisors, h is injective. Because R is
assumed finite-dimensional over K, it is also surjective. In particular, there exists
b 2 R such that ab D 1.
˜
Remark. It can be proved in an analogous way that an integral domain that has
finite cardinality is a field.
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˜
The minimal polynomial
17
3. Let E=K be a field extension, and let ˛ 2 E be algebraic over K. Consider
on the K-vector space K.˛/ the endomorphism h defined by multiplication by ˛.
The minimal polynomial of h is called the minimal polynomial of ˛ over K, and we
denote it by
MiPoK .˛/:
This is the lowest-degree normalized polynomial in KŒX that has ˛ as a zero. (That
there can be only one such polynomial is clear: if f; g are both normalized and of
degree n, the degree of f g is less than n.) The degree of f D MiPo˛ .K/ is also
called the degree of ˛ over K, and is denoted by Œ˛ W K.
Example. Consider E D ރ, K D ޑand ˛ D e 2 i=3 . Then ˛ is a root of X 3 1. But
X 3 1 D .X 1/g.X /, with g.X / D X 2 C X C 1; since Ô 1, we have g./ D 0.
Let f D MiPoK .˛/; we claim that f D g. Otherwise necessarily deg f < deg g, so
f could only be of the form f .X / D X ˛, which is impossible since ˛ … ޒ.
F3. Let E=K be a field extension and let ˛ 2 E be algebraic over K, of degree
n WD Œ˛ W K. The elements
1; ˛; ˛ 2 ; : : : ; ˛ n
(6)
1
of E form a basis of K.˛/ over K. In particular,
(7)
K.˛/ W K D Œ˛ W K D deg MiPoK .˛/:
Proof. Let f .X / D X n C
know that
C a1 X C a0 the minimal polynomial of ˛ over K. We
K.˛/ W K Ä nI
see (5) in the proof of F1. There remains to show that 1; ˛; ˛ 2 ; : : : ; ˛ n
independent over K. Suppose there is a relation
(8)
n 1
X
ci ˛ i D 0
1
are linearly
with ci 2 K:
iD0
P
Set g.X / WD niD01 ci X i . If some ci in (8) were nonzero, g.X / would be a nonzero
polynomial in KŒX of degree less than n and vanishing at ˛. Contradiction!
˜
4. Let E=K be a field extension and assume ˛ 2 E is algebraic over K. Is it the
case that any ˇ 2 K.˛/ is also algebraic over K?
Definition. An extension E=K is called algebraic if every element of E is algebraic
over K. An extension E=K is called finite if E W K < 1.
Remarks. ޒ= ރis a finite extension, since ރW ޒD 2. The extension ޑ= ޒis not
algebraic; see Remark (b) in Section 2.1.
An extension E=K is called transcendental if it is not algebraic.
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18
2 Algebraic Extensions
F4. If an extension E=K is finite, it is also algebraic; for each ˇ 2 E the degree
Œˇ W K is a divisor of E W K.
Proof. Let E=K be finite of degree n. Given ˇ 2 E, the n C 1 elements 1; ˇ;
ˇ 2 ; : : : ; ˇ n of the n-dimensional K-vector space E are linearly dependent. Therefore
there exist a0 ; a1 ; : : : ; an 2 K, not all zero, such that
a0 1 C a1 ˇ C
C an ˇ n D 0:
Thus ˇ is algebraic over K. By F3, Œˇ W K D K.ˇ/ W K, and K.ˇ/ W K is a divisor
of E W K by the degree formula (Chapter 1, F7).
˜
We now can easily answer in the affirmative the question asked at the beginning
of this section.
F5. Let E=K be a field extension. If ˛ 2 E is algebraic over K, the extension
K.˛/=K is algebraic.
Proof. If ˛ is algebraic over K, we know from F1 that K.˛/=K is finite. But every
finite field extension is algebraic, by F4.
˜
Together, F1 and F4 afford the following criterion:
F6. Let E=K be a field extension. An element ˛ of E is algebraic over K if and only
if K.˛/=K is finite.
Now it is a cinch to prove Theorem 1, which we can reformulate as follows:
Theorem 1. Let M be a subset of ރcontaining 0 and 1. Let K D ޑ.M [ M /. The
field extension M =K is algebraic.
Proof. Take z 2 M . From F9 of Chapter 1 we know that K.z/ W K < 1. Then F6
says that z is algebraic over K.
˜
Remark. The converse of F4 is not true: Not every algebraic extension is finite.
This will soon become obvious. In fact a counterexample comes up naturally in our
context: If E D f0; 1g is the field of all numbers constructible from f0; 1g with
ruler and compass, the field extension E= ޑis algebraic but not finite. (With what
we know so far this is not very easy to prove, but it’s worth thinking about; see §2.5
in the Appendix.)
Among algebraic extensions, finite extensions can be characterized thus:
F7. Let E=K be a field extension. The following conditions are equivalent:
(i) There are elements ˛1 ; : : : ; ˛m of E, finite in number and algebraic over K,
such that E D K.˛1 ; : : : ; ˛m /.
(ii) E=K is finite.
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Properties of algebraic extensions
19
Proof. (ii) ) (i) is clear; all we need to do is choose a basis ˛1 ; : : : ; ˛m for E=K.
Then we actually have E D K˛1 C C K˛m , and by F4 all the ˛i are algebraic
over K.
To show (i) ) (ii) we use induction over m. For m D 0 there is nothing to
prove. Assume that (i) holds for some m 1 and set
K 0 D K.˛1 ; : : : ; ˛m
1 /:
Then E D K 0 .˛m /. Since ˛m is algebraic over K, it is a fortiori algebraic over the
larger field K 0 . By F1 this implies E W K 0 < 1. But by the induction hypothesis,
K 0=K is finite. The degree formula (Chapter 1, F7) then implies that E=K is finite.
˜
5. Let E=K be a field extension. A subfield L of E containing K is called an
intermediate field of the extension E=K.
F8. Let E=K be a field extension. The subset
F D f˛ 2 E j ˛ is algebraic over Kg
is an intermediate field of E=K. It is called the algebraic closure of K in E. In
particular, the set of all algebraic numbers is a subfield of ރ.
Proof. Take ˛; ˇ 2 F . Consider the subfield K.˛; ˇ/ of E. By F7 the extension
K.˛; ˇ/=K is finite (prove this again for practice). Now apply F4; all elements of
K.˛; ˇ/ are algebraic over K, so
K.˛; ˇ/ Â F:
The elements ˛ Cˇ, ˛ ˇ, ˛ˇ and 1= (if Ô 0) lie in K.; /, and thus also in
F . So F really is a subfield of E. Clearly K Â F , since any ˛ 2 K is a zero of
a polynomial X ˛ 2 KŒX and therefore algebraic over K. This completes the
proof.
˜
This proof qualifies as easy, but it’s only easy because we have the right notions
at our disposal. Otherwise, would you be able to write down, at the drop of a hat, a
nontrivial rational polynomial that vanishes at the sum of two numbers, given only
rational polynomials vanishing at one and the other number respectively?
F9 (Transitivity of algebraicness). Let L be an intermediate field of the extension
E=K. If E=L and L=K are algebraic, so is E=K (and vice versa).
Proof. Take ˇ 2 E. By assumption ˇ is algebraic over L. Let ˛0 ; ˛1 ; : : : ; ˛n 1
be the coefficients of MiPoL .ˇ/; then ˇ is also algebraic over the subfield F WD
K.˛0 ; ˛1 ; : : : ; ˛n 1 /. By assumption all the ˛i are algebraic over K. Therefore we
can apply F7 to conclude that F W K is finite. But F.ˇ/ W F is also finite, by F6;
therefore the degree formula gives
F.ˇ/ W K < 1:
Using F4 we see in particular that ˇ is algebraic over K.
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˜
20
2 Algebraic Extensions
F10. Let E=K be a field extension and A a subset of E. If all elements of A are
algebraic over K, the extension K.A/=K is algebraic.
Proof. Clearly K.A/ is the union of all subfields of the form K.M /, where M
ranges over finite subsets of A. By F7, each K.M /=K is finite and therefore also
algebraic. Thus K.A/ contains only elements algebraic over K. (Of course F10
also follows directly from F8.)
˜
F11. Let E=K be a field extension, and L1 ; L2 intermediate fields of E=K. The field
(9)
L1 L2 WD L1 .L2 / D L2 .L1 /
is called the composite of L1 and L2 in E.
(a) If L1 =K is algebraic, so is L1 L2 =L2 .
(b) If L1 =K is finite, so is L1 L2 =L2 ; moreover L1 L2 W L2 Ä L1 W K.
(c) If L1 =K and L2 =K are algebraic, so is L1 L2 =K.
(d) If L1 =K and L2 =K are finite, so is L1 L2 =K; if , moreover, the extension
degrees n1 D L1 W K and n2 D L2 W K are relatively prime, we have L1 L2 W K D
n1 n2 .
Proof. Part (a) follows from F10, taking (9) into account. Part (c) therefore also
follows, thanks to F9. Let L1 =K and L2 =K be finite. Assuming (b) already proved,
we see from the degree formula that
(10)
L1 L2 W K D .L1 L2 W L2 /.L2 W K/ Ä .L1 W K/.L2 W K/;
which is the first part of (d). Again from the degree formula we obtain that L1 L2 W K
is divisible by n1 and by n2 . If n1 ; n2 are relatively prime, L1 L2 W K is divisible by
n1 n2 , which together with (10) gives the second part of (d).
There remains to prove (b). Consider the set R of all finite sums of products ab
with a 2 L1 ; b 2 L2 . Clearly R is a subring of E containing L1 and L2 . It is also
clear that any basis of L1 =K generates R as an L2 -vector space R, so in particular
R W L2 Ä L1 W K. If L1 W K < 1, this implies that R is a field (see F2). It follows
that R D L1 L2 , which concludes the proof.
˜
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